Problem: Simplify the following expression and state the condition under which the simplification is valid. $p = \dfrac{-8r^2 + 104r - 320}{3r^2 - 27r + 24}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ p = \dfrac {-8(r^2 - 13r + 40)} {3(r^2 - 9r + 8)} $ $ p = -\dfrac{8}{3} \cdot \dfrac{r^2 - 13r + 40}{r^2 - 9r + 8} $ Next factor the numerator and denominator. $ p = - \dfrac{8}{3} \cdot \dfrac{(r - 8)(r - 5)}{(r - 8)(r - 1)}$ Assuming $r \neq 8$ , we can cancel the $r - 8$ $ p = - \dfrac{8}{3} \cdot \dfrac{r - 5}{r - 1}$ Therefore: $ p = \dfrac{ -8(r - 5)}{ 3(r - 1)}$, $r \neq 8$